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Proof step 12 from proof div-mult-two-nls-01 (previous step)

(2 * i0) * j + 2 * j0 = i1 * j + j1, j1 < j, j0 < j, 1 + 2 * i0 < i1, 0 < j, 0 ≤ j0, 0 ≤ i0, 0 ≤ j1, 0 ≤ i1 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-mult inst: name: mods-eq seq: ⊦ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ 0 ≤ j1 ∧ j0 < j ∧ j1 < j → (i * j + j0 = i0 * j + j1 ↔ i = i0 ∧ j0 = j1) subst: [j1, i0, j0, j, i] → [j1, i1, 2 * j0, j, 2 * i0] flags: rewrite

Additional information

Used simplifier rules: zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%mods-ls-04: ⊦ j0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 ≤ j0 → (i * j + j0 < i0 * j ↔ i < i0); from specification int-mult%irref: ⊦ ¬ i < i; from specification int-basic1%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: j1, i1, 2 * j0, j, 2 * i0